Let $a, b, c$ be the lengths of sides of triangle $\mathrm{ABC}$ such that…

Let $a, b, c$ be the lengths of sides of triangle $\mathrm{ABC}$ such that $\frac{\mathrm{a}+\mathrm{b}}{7}=\frac{\mathrm{b}+\mathrm{c}}{8}=\frac{\mathrm{c}+\mathrm{a}}{9}=\mathrm{k}$. Then $\frac{(\mathrm{A}(\triangle \mathrm{ABC}))^2}{\mathrm{k}^4}=$
  1. 36
  2. 32
  3. 38
  4. 40

Solution

$\begin{aligned} & \Rightarrow \text { In } \triangle \mathrm{ABC} \text {, } \\ & \frac{\mathrm{a}+\mathrm{b}}{7}=\frac{\mathrm{b}+\mathrm{c}}{8}=\frac{\mathrm{c}+\mathrm{a}}{9} \Rightarrow \mathrm{k} \\ & \therefore \quad \mathrm{a}+\mathrm{b}=7 \mathrm{k} ...(i)\\ & \mathrm{b}+\mathrm{c}=8 \mathrm{k} ... (ii) \\ & \mathrm{c}+\mathrm{a}=9 \mathrm{k} ... (iii)\\ & \end{aligned}$ Adding above equations, $\begin{aligned} & 2 a+2 b+2 c=24 k \\ & a+b+c=12 k ... (iv) \end{aligned}$ Solving equations (i), (ii), (iii), (iv) We get, $\begin{array}{ll} & \mathrm{c}=5 \mathrm{k}, \mathrm{a}=4 \mathrm{k}, \mathrm{b}=3 \mathrm{k} \\ \therefore \quad & \mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2 \end{array}$ $\therefore \quad \triangle \mathrm{ABC}$ is right angled triangle $\therefore \quad \angle \mathrm{C}=90^{\circ}$ $\begin{aligned} & \text { Area of } \begin{aligned} \triangle \mathrm{ABC} & =\frac{1}{2} \mathrm{ab} \sin \mathrm{C} \\ & =\frac{1}{2} \mathrm{ab} \sin 90 \\ & =\frac{1}{2} \times 4 \mathrm{k} \times 3 \mathrm{k} \\ & =6 \mathrm{k}^2\end{aligned} \\ & \therefore \quad \text { Now, } \frac{[\mathrm{A}(\Delta \mathrm{ABC})]^2}{\mathrm{k}^4}=\frac{\left(6 \mathrm{k}^2\right)^2}{\mathrm{k}^4}=\frac{36 \mathrm{k}^4}{\mathrm{k}^4}=36\end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 2)

Practice more Properties of Triangles questions on Aicharya